# Find the area of the following region y=sin x and y=sin 2x from x=0 x=pi/2

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### 1 Answer

The formula for area between two curves is:

`A = int_a^b (y_U - y_L) dx`

The integral limits a and b are the intersection points of the two equation. To determine, use elimination method of system of equations.

Subtract the two equations to eliminate y variable.

`y = sin x`

`(-) ` `y = sin 2x`

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`0 = sin x - sin 2x`

To simplify, replace sin 2x with 2sinx cos x. Note that from double angle identity sin 2x = 2sin x cos x .

` 0 = sinx - 2sinx cosx`

`0 = sin x (1-2cos x)`

Then, set each factor to zero and solve for x.

`sinx = 0 ` and `1 - 2cos x = 0`

`x = 0 ` ` 2cos x = 1`

`cos x = 1/2`

`x = pi/3`

Substitute the values of x to either original equations.

`x = 0` , `y = sinx =sin 0 = 0`

`x = pi/3` , `y = sin x = sin pi/3 = sqrt3/2`

Hence at the interval `0lt=xlt=pi/2` , the intersection points are `(0,0)` and `(pi/3, sqrt3/2)` .

Then plot the two equations.

*(The graph of y=sin x is in green. While y=sin 2x is the blue one.)*

Base on the graph, at interval `0lt=xlt=pi/2` , `y_U=sin 2x` and `y_L=sinx` .

`A = int_0^(pi/3) (y_U-y_L)dx = int_0^(pi/3) (sin2x - sinx)dx`

`A = int_0^(pi/3)sin2x dx - int_0^(pi/3) sinxdx = -(cos 2x)/2 |_0^(pi/3) - (-cos x)|_0^(pi/3) `

`A = -cos(2x)/2 | _0^(pi/3) + cos x |_0^(pi/3)`

`A = [-cos((2pi)/3)/2 - (-cos (2*0)/2)] + [ cos (pi/3) - cos 0]`

`A = -cos((2pi)/3)/2 + cos 0 /2 + cos(pi/3) - cos 0`

`A= - (-1/2)/2 + 1/2 + 1/2 - 1 = 1/4`

**Hence, area bounded by `y=sinx ` and `y=sin 2x` at the interval `0lt=xltpi/2` is `1/4` square units.**